Shell脚本中判断输入变量或者参数是否为空的方法

1.判断变量

read -p "input a word :" word
if  [ ! -n "$word" ] ;then
    echo "you have not input a word!"
else
    echo "the word you input is $word"
fi
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2.判断输入参数

#!/bin/bash
if [ ! -n "$1" ] ;then
    echo "you have not input a word!"
else
    echo "the word you input is $1"
fi
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以下未验证。

3. 直接通过变量判断

如下所示:得到的结果为: IS NULL
#!/bin/sh
para1=
if [ ! $para1 ]; then
  echo "IS NULL"
else
  echo "NOT NULL"
fi
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4. 使用test判断

得到的结果就是: dmin is not set!
#!/bin/sh
dmin=
if test -z "$dmin"
then
  echo "dmin is not set!"
else 
  echo "dmin is set !"
fi
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5. 使用""判断

#!/bin/sh
dmin=
if [ "$dmin" = "" ]
then
  echo "dmin is not set!"
else 
  echo "dmin is set !"
fi
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下面是我在某项目中写的一点脚本代码， 用在系统启动时:
#! /bin/bash

echo "Input Param Is [$1]"

if [ ! -n "$1" ] ;then
 echo "you have not input a null word!"
 ./app1;./app12;./app123
elif [ $1 -eq 2 ];then
 ./app12;./app123
elif [ $1 -eq 90 ];then
 echo "yy";
fi
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